Proof of non-existence of perfect cuboids

Let $a$, $b$, $c$, $d$, $e$, $f$ and $g$ be positive integers, $a$, $b$ and $c$ be the edge lengths, $d$, $e$ and $f$ be the diagonal lengths and $g$ be the space diagonal length. We consider that $a$, $b$ and $c$ have no common prime factors. If a perfect cuboid exists, the following equations hold.
$$a^2+b^2=d^2$$
$$b^2+c^2=e^2$$
$$c^2+a^2=f^2$$
$$a^2+b^2+c^2=g^2$$
We regard that $a$ is odd and $b$ and $c$ are even because $a$, $b$ and $c$ are not all even and if two edges are odd, then the diagonal is not a square number. And $d$ and $f$ are odd and $e$ is even. Let $i$, $k_i$, $m_i$ and $n_i$ be positive inetgers and $1≦i≦6$ hods. We consider that $m_i$ and $n_i$ are relatively prime and the odd/even number of $m_i$ and $n_i$ are different. By the form of Pythagorean triple, the following equations hold.
$$a=(m_1^2-n_1^2)k_1=(m_2^2-n_2^2)k_2=(m_5^2-n_5^2)k_5$$
$$b=2k_2m_2n_2=(m_3^2-n_3^2)k_3=2k_6m_6n_6$$
$$c=2k_3m_3n_3=2k_1m_1n_1=2k_4m_4n_4$$
$$d=(m_2^2+n_2^2)k_2=(m_4^2-n_4^2)k_4$$
$$e=(m_3^2+n_3^2)k_3=2k_5m_5n_5$$
$$f=(m_1^2+n_1^2)k_1=(m_6^2-n_6^2)k_6$$
$$g=(m_4^2+n_4^2)k_4=(m_5^2+n_5^2)k_5=(m_6^2+n_6^2)k_6\ ...(1)$$
$k_3$ is even and the others $k_i$ are all odd. We consider that $k_4$, $k_5$ and $k_6$ are mutually prime to each other since if two of the diagonal lengths have a common prime factor, then the length of the space diagonal and the length of the other diagonal will also have the prime factor, three edge lengths will have it too and this contradicts the condition that $a$, $b$ and $c$ have no common prime factors. Let $r$ be an odd positive integer. Since $g$ has factors $k_4$, $k_5$ and $k_6$,
$$g^2=((m_5^2-n_5^2)k_5)^2+(2k_6m_6n_6)^2+(2k_4m_4n_4)^2=(rk_4k_5k_6)^2\ ...(2)$$
holds. Let $s$ and $u$ be odd positive integers and $t$ be an even positive integer. By this equation, the following equations hold.
$$((m_5^2-n_5^2)k_5)^2+(2k_6m_6n_6)^2=sk_4^2$$
$$(2k_6m_6n_6)^2+(2k_4m_4n_4)^2=tk_5^2$$
$$(2k_4m_4n_4)^2+((m_5^2-n_5^2)k_5)^2=uk_6^2$$
By these equations,
$$sk_4^2+tk_5^2+uk_6^2=2(k_4k_5k_6)^2$$
holds. Let $v$ and $w$ be odd positive integers and $x$ be an even positive integer. By this equation, the following equations hold.
$$sk_4^2+tk_5^2=vk_6^2$$
$$tk_5^2+uk_6^2=wk_4^2$$
$$uk_6^2+sk_4^2=xk_5^2$$
By these equations,
$$wk_4^2+xk_5^2+vk_6^2=(2rk_4k_5k_6)^2\ ...(3)$$
holds. Comparing the equation (3) with the equation (2), we find that the following equations hold since the equations (2) and (3) must hold for any $k_4$, $k_5$ and $k_6$ satisfying the equation (1).
$$v=(4m_6n_6)^2$$
$$w=(4m_4n_4)^2$$
$$x=(2(m_5^2-n_5^2))^2$$
It becomes a contradiction since $v$ and $w$ are odd. From the above, there are no perfect cuboids. (Q.E.D.)